[R-meta] Meta-Analysis: Proportion in overall survival rate

Dr. Gerta Rücker ruecker @end|ng |rom |mb|@un|-|re|burg@de
Wed May 27 22:16:36 CEST 2020

Dear Nelly,

I forgot to add that my suggestion was thought as an approach when 
mixing directly reported proportions and proportions estimated from 
Kaplan-Meier curves (with the caveat I mentioned before in my first post).

However, if you really restrict your analysis to those studies that 
offer a Kaplan-Meier estimator, you may not need this proportion 
approach at all, if standard errors/confidence intervals for the 
Kaplan-Meier curves at time point 2 years are available. Then you could 
take these standard errors as se in the rma call.



Am 27.05.2020 um 22:05 schrieb Dr. Gerta Rücker:
> Dear Nelly, dear all,
> I have another problem with this approach, and this is the 
> denominator, "n". You insert n_t for n, but n_t = n_0*S(t) estimates 
> the number of patients surviving at time t which is the numerator of 
> the ratio. The denominator is n_0, not n_t. And n_0 should be used as n.
> (Suppose a study without any censoring. Then the Kaplan-Meier 
> estimator in t gives exactly p = n_t/n_0 which is the same as if the 
> study reports simply the 2-year survival proportion relative to the 
> sample size, n_0.)
> Thus the n in your calculation should be n_0, irrespectively of the 
> transformation method (I agree with Wolfgang that the logit 
> transformation is preferable).
> I don't see a principle problem with this approach, because n_t is an 
> unbiased estimate of the numerator and sample size n_0 is - for sure - 
> the denominator.
> Best,
> Gerta
> Am 27.05.2020 um 21:07 schrieb ne gic:
>> Many thanks for the insights, WoIfgang!
>> I concur, the proportion is likely not from a binomial distribution, 
>> so I
>> take your advice.
>> Sincerely,
>> nelly
>> On Wed, May 27, 2020 at 8:17 PM Viechtbauer, Wolfgang (SP) <
>> wolfgang.viechtbauer using maastrichtuniversity.nl> wrote:
>>> Dear Nelly,
>>> Your equation for the SE assumes that the p behaves like a 'regular'
>>> proportion computed from a binomial distribution. I am not sure if 
>>> this is
>>> correct when using the Kaplan-Meier estimator to derive such a 
>>> proportion.
>>> As far as your input to rma() is concerned - that is correct. 
>>> However, I
>>> would consider not meta-analyzing the proportions directly, but doing a
>>> logit transformation on p, so using qlogis(p) for yi and sqrt(1/(p*n) +
>>> 1/((1-p)*n)) for the SE.
>>> Best,
>>> Wolfgang
>>>> -----Original Message-----
>>>> From: R-sig-meta-analysis [mailto:
>>> r-sig-meta-analysis-bounces using r-project.org]
>>>> On Behalf Of ne gic
>>>> Sent: Wednesday, 27 May, 2020 20:02
>>>> To: Dr. Gerta Rücker
>>>> Cc: r-sig-meta-analysis using r-project.org
>>>> Subject: Re: [R-meta] Meta-Analysis: Proportion in overall survival 
>>>> rate
>>>> Dear Michael, Gerta and List,
>>>> I would like to cross-check with you what I have done.
>>>> I have restricted myself to Kaplan-Meier studies which gave the 
>>>> number at
>>>> risk at 2 years, and also n_0 at baseline.
>>>> I then estimated the absolute number of those surviving as *n_t *=
>>> n_0*S(t)
>>>> following Gerta's idea. I took the reported proportions at 2 years to
>>>> represent the S(t).
>>>> I calculated the standard error (SE) using the formula: *se *= square
>>> root (
>>>> *p*(1-*p*)/n). Where *p* = proportion at 2 years i.e. S(t)
>>>> , n = *n_t*, the estimated number of of those surviving.
>>>> I then used the random effects model in metafor as follows:
>>>> rma(yi = *p*, sei = *se*, data=mydata, method="REML")
>>>> The resulting estimate seems reasonable to me. But I want to 
>>>> confirm with
>>>> you if this is the way one would input SE and the proportion to the
>>>> function.
>>>> Welcome any comments.
>>>> Sincerely,
>>>> nelly
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